Graph of fourth degree equation? - Answers

Every polynomial defines a function. A function of fourth degrees have a form such as,f(x) = ax^4 + bx^3 + cx^2 + dx + e = 0, where a is different than 0For example, let's graph f(x) = x^4 - 2x^2 +1 (Here the terms with x^3, and x, are missed. I gave this example because students need to know that a fourth degree equation exists even though is different from the general formula. Since it has the power of 4, it still is a fourth degrees equation)Solution:Step 1. Use the Leading Coefficient Test to determine the graph's end behavior.Because the degrees of f(x) = 1x^4 - 2x^2 + 1 is even (n = 4) and the leading coefficient , 1, is positive, the graph rises to the left and rises to the right.Step 2. Find x-intercepts by setting f(x) = 0 and solve the resulting polynomial equation.x^4 - 2x^2 + 1 = 0 factor; use the formula (a - b)^2 = (a - b)(a- b)((x^2 -1)(x^2 -1) = 0 factor completely; use the formula (a^2 - b^2) = ( a - b)(a + b);(x - 1)(x +1)(x - 1)(x + 1) = 0(x - 1)(x - 1)(x +1)(x +1) = 0(x - 1)^2(x + 1)^2 = 0 solve for x;x = 1 or x = -1 you have double roots here; so,the graph touches x-axis at -1 and 1 and turns around.Step 3. Find the y-intercept by computing f(0).f(x) = x^4 - 2x^2 + 1f(0) = 0^4 - 2(0)^2 + 1 = 1There is a y-intercept ay i, so the graph passes through (0, 1).Step 4. use possible symmetry to help draw the graph. It will be y-axis symmetry. Let,s verify this by finding f(-x).f(x) = x^4 - 2x^2 + 1f(-x) = (-x)^4 - 2(-x)^2 +1f(-x) = x^4 - 2x^2 + 1Because f(-x) = f(x), the graph of f is symmetric with respect to the y-axis.You are able to graph now.Step 5. Use the fact that maximum number of turning points of the graph is n -1 to check whether it is drawing correctly. because you are graphing f(x) = x^4 - 2x^2 + 1, with n = 4, the maximum number of turning points is 4 -1 or 3. So, your graph will have 3 turning points, if it has more, you have violated the maximum number possible.Verify this observation.